Programming challenge: rotating a matrix 90 degrees in place

RotatingMatrix90DegreesInPlace.java

// Programming challenge: rotating a matrix 90 degrees in place
// Original post: https://blog.svpino.com/2015/05/10/programming-challenge-rotating-a-matrix-90-degrees-in-place

public class RotatingMatrix90DegreesInPlace {

    private static int[][] matrix = { 
        { 1, 2, 3, 4 }, 
        { 5, 6, 7, 8 }, 
        { 9, 10, 11, 12 }, 
        { 13, 14, 15, 16 } 
    };

    private final static int N = 4;

    private static void print() {
        for (int i = 0; i < N; i++) {
            for (int j = 0; j < N; j++) {
                System.out.print("\t" + matrix[i][j]);
            }
            System.out.println();
        }
        System.out.println();
    }

    public static void main(String[] args) {
        print();

        for (int ring = 0; ring < N / 2; ring++) {
            int farthest = N - ring - 1;

            for (int i = ring; i < farthest; i++) {
                int temp = matrix[ring][i];
                matrix[ring][i] = matrix[farthest - i + ring][ring];
                matrix[farthest - i + ring][ring] = 
                    matrix[farthest][farthest - i + ring];
                matrix[farthest][farthest - i + ring] = 
                    matrix[i][farthest];
                matrix[i][farthest] = temp;
            }
        }

        print();
    }
}

You can simply do two pass process to the matrix: first transpose, and then flip horizontally, that means the subscript changes as (i, j)->(j, i)->(j, n-1-i). Code following:

public class RotatingMatrix90DegreesInPlace {

private static int[][] matrix = { 
    { 1, 2, 3, 4, }, 
    { 5, 6, 7, 8 }, 
    { 9, 10, 11, 12 }, 
    { 13, 14, 15, 16 } 
    // {1, 2, 3, 4, 5,},
    // {6, 7, 8, 9, 10},
    // {11, 12, 13, 14, 15,},
    // {16, 17, 18, 19, 20},
    // {21, 22, 23, 24, 25}
};

private final static int N = matrix.length;

private static void print() {
    for (int i = 0; i < N; i++) {
        for (int j = 0; j < N; j++) {
            System.out.print("\t" + matrix[i][j]);
        }
        System.out.println();
    }
    System.out.println();
}

public static void main(String[] args) {
    print();

    for (int i = 0; i < N; ++i) {
        for (int j = i + 1; j < N; ++j) {
            int tmp = matrix[i][j];
            matrix[i][j] = matrix[j][i];
            matrix[j][i] = tmp;
        }
    }

    for (int i = 0; i < N; ++i) {
        for (int j = 0; j < N - 1 - j; ++j) {
            int tmp = matrix[i][j];
            matrix[i][j] = matrix[i][N - 1 - j];
            matrix[i][N - 1 - j] = tmp;
        }
    }

    print();
}

}

You can simply do two pass process to the matrix: first transpose, and then flip horizontally, that means the subscript changes as (i, j)->(j, i)->(j, n-1-i).

I was working with matrices the other day and came up with the same solution, and then thought back to the post. I was writing Haskell, where this literally looks like this:

import Data.List (transpose)

cw :: [[a]] -> [[a]]
cw = (map reverse) . transpose

ccw :: [[a]] -> [[a]]
ccw = reverse . transpose

Note that by reversing each of the columns, we get a clockwise rotation and when reversing the columns themselves, we get a counter-clockwise rotation. Obviously, because this is Haskell, there is no in-place transformation, because data is immutable.

I did it in two steps,
First build the Transpose
Then Swap columns
I feel this solution is easier to implement , I dont know. :)
https://gist.github.com/Qassas/1336b9c6d27fcd933d4bea8f76e40827#file-programming-challenge-rotating-a-matrix-90-degrees-in-place-cpp